1/（sinx的n次方）的不定积分肿么求呀……只要递推式就行了
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I = ∫1/[sinx]^n dx = ∫[cscx]^n dx =∫[cscx]^[n-2] * csc?x dx = -∫[cscx]^[n-2] dcotx = -cotx * [cscx]^[n-2] + ∫cotx d[cscx]^[n-2] = -cotx * [cscx]^[n-2] + [n-2]∫cotx * -[cscx]^[n-3] * cscx*cotx dx = -cotx * [cscx]^[n-2] - [n-2]∫cot?x * [cscx]^[n-2] dx = -cotx * [cscx]^[n-2] - [n-2]∫(csc?x-1) * [cscx]^[n-2] dx = -cotx * [cscx]^[n-2] - [n-2]*I + [n-2]∫[cscx]^[n-2] dx (1+n-2)*I = -cotx * [cscx]^[n-2] + [n-2]∫[cscx]^[n-2] dx I = -[cotx * (cscx)^(n-2)]/(n-1) + (n-2)/(n-1) * ∫(cscx)^(n-2) dx

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